cf580D. Kefa and Dishes(状压dp)
题意$n$个食物，每个食物有一个满意度，从中选出$m$个，使得满意度最大 同时有$k$个关系：若$x_i$在$y_i$之前吃，则会获得$C_i$的代价 Sol官方题解是$O(2^n n^2)$的，不过我没发现状态之间的联系，就写了一个$O(2^n n^3)$的，不过还是水过去了。 $f[i][j][sta]$表示现在已经放了$i$个，本轮要放第$j$个，状态为$sta$ 转移的时候枚举一下上一个放了什么 /* */ #include #include #include #include #include #include #include #include #include #include #include #define Pair pair #define MP(x,y) make_pair(x,y) #define fi first #define se second #define int long long #define LL long long #define rg register #define sc(x) scanf("%d",&x); #define pt(x) printf("%d ",x); #define db(x) double x #define rep(x) for(int i = 1; i <= x; i++) //#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf,1,1<<22,stdin),p1 == p2) ? EOF : *p1++) //char buf[(1 << 22)],*p1 = buf,*p2 = buf; char obuf[1<<24],*O = obuf; #define OS *O++ = ' '; using namespace std; using namespace __gnu_pbds; const int MAXN = 1e6 + 10,INF = 1e9 + 10,mod = 1e9 + 7; const double eps = 1e-9; inline int read() { char c = getchar(); int x = 0,f = 1; while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();} while(c >= '0' && c <= '9') x = x * 10 + c - '0',c = getchar(); return x * f; } void print(int x) { if(x > 9) print(x / 10); *O++ = x % 10 + '0'; } int N,M,K; int f[2][19][262145]; int lk[19][19],a[MAXN]; main() { N = read(); M = read(); K = read(); for(int i = 1; i <= N; i++) a[i] = read(); for(int i = 1; i <= K; i++) { int x = read(),y = read(),z = read(); lk[x][y] = z; } int lim = (1 << N) - 1,o = 0; for(int i = 1; i <= M; i++) { o ^= 1; for(int sta = 0; sta <= lim; sta++) { if((__builtin_popcount(sta)) != i) continue; for(int j = 1; j <= N; j++) {//???????? if(!(sta & (1 << j - 1))) continue; for(int k = 1; k <= N; k++) {//???????? if(sta & (1 << k - 1)) { /*if(o == 0 && j == 2 && k == 1) { puts("GG"); }*/ f[o][j][sta] = max(f[o][j][sta],f[o ^ 1][k][sta ^ (1 << j - 1)] + lk[k][j]); } } } } } // printf("%dn",f[o][2][lim]); int ans = 0; for(int i = 1; i <= N; i++) { for(int sta = 0; sta <= lim; sta++) { if((__builtin_popcount(sta)) != M) continue; if(!(sta & (1 << i - 1))) continue; int now = f[o][i][sta]; for(int j = 1; j <= N; j++) if(sta & (1 << j - 1)) now += a[j]; ans = max(ans,now); } } cout << ans; return 0; } /* 2 2 1 1 1 2 1 1 */ （编辑：李大同） 【声明】本站内容均来自网络，其相关言论仅代表作者个人观点，不代表本站立场。若无意侵犯到您的权利，请及时与联系站长删除相关内容! |